Lightweight Electric/Hybrid Vehicle Design
Lightweight Electric/Hybrid Vehicle Design
Lightweight Electric/Hybrid Vehicle Design
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<strong>Design</strong> for optimum body-structural and running-gear performance efficiency 213<br />
known as tension coefficients is applied. This is based on the fact that proportionality exists between<br />
both resolved components of length and of force. A tension coefficient<br />
S/L = F x /X = F y /Y = F z /Z<br />
for member force S, in a length L, has projecting force vectors and length components, F x , F y , F z<br />
and X,Y,Z on perpendicular coordinate axes.<br />
The view at (d) shows one bay of a space truss, that might be an extension used to support an<br />
engine/gearbox unit behind a monocoque bodyshell structure. For analysis, the frame is assumed<br />
to have rigid plates at b 1,2,3,4 and c 1,2,3,4 which offer no force reaction perpendicular to their own<br />
planes (zero axial warping constraint). When considering the torsional load case, for the bay, b 1 c 1<br />
and b 4 c 4 member forces are zero by resolving at b 1 and c 4 , observing zero axial constraint of the<br />
truss. Remaining members (the envelope) have equal force components F 2 at the bulkheads and<br />
tension coefficients for all of them are F z /Z. This common value can be determined by projecting<br />
envelope member forces onto the bulkhead planes, as shown. By taking moments about any axis<br />
O, for any one member (c 3 b 4 , say, with projected length d), contribution to torque reaction is = rtd<br />
(half the area of the triangle formed by joining O to c 3 and b 4 ). Thus total torque is 2tA, by summation.<br />
8.4.1 STIFF-JOINTED FRAMES<br />
Where structural members are curved or cranked over wheel arches or drive shaft tunnels the unit<br />
load method of analysis applied to stiff-jointed frames is particularly useful, Fig. 8.8. It is best<br />
understood by considering a small elastic element in a curved beam (a) which is otherwise assumed<br />
to be rigid. An imaginary load w, at P, is then considered to cause vertical displacement Δ; the<br />
beam deflects under the influence of the bending moment and it can be shown that Δ = ∫ Mm/<br />
EI . ds where m is bending moment due to imaginary unit load at P and M due to the real external<br />
load system. An example might be a car floor transverse crossbearer having a ‘tunnel’ portion for<br />
exhaust-system and prop-shaft clearance, (b). To find the downward deflection at B, the imaginary<br />
unit load is applied at that point and it develops reactions of two-thirds at A and one-third at D.<br />
The bending moments for the real and imaginary loads are shown at (c) which also illustrates the<br />
subsequent calculation to obtain the deflection due to the driver and seat represented by distributed<br />
load w over length l. Another example is the part of the frame at (d) subject to twisting and<br />
bending, deflection due to twist = ∫ Tt dx/GJ. This deflection would then be added to that due to<br />
bending, obtained from the formula given in the earlier section. Usually the deflection due to axial<br />
straining of the elements is so small that it can be discounted.<br />
A good example of a frame subject to twisting and bending is the portal shape frame at (e)<br />
having loading and support, in the horizontal plane, indicated. If it is required to find the vertical<br />
deflection at D for the load P applied at B, equating vertical forces and moments is first carried out<br />
to find the reactions at the supports. Next step is to break the frame up into its elements and<br />
determine the bending moments in each – ensuring compatibility of these, and that associated<br />
loads are ‘transferred’ across the artificially broken joints. Integrations have to be carried out for<br />
the three deflection modes as follows: ∫ Sb . ds/AE + ∫ Mm . ds/EI + ∫ Ff . ds/AG to obtain the<br />
combined deflection. Both vertical and horizontal components can be obtained by applying vertical<br />
and horizontal imaginary unit loads, in turn, at the point where the deflection is to be determined.<br />
For the example shown in the figure, elastic and shear moduli, E and G, are 200 and 80 GN/m 2<br />
respectively. S is axial load and F the shear load, the latter being negative in compression. The<br />
values of these, with the bending moments, are as shown at (f right). These can readily be substituted<br />
in the expression (f left) to determine deflections as a factor of P.